Question

On a 20 km tunnel, connecting two cities A and B, there are three gutters (1, 2, and 3). The distance between gutters 1 and 2 is half the distance between gutters 2 and 3. The distance from city A to its nearest gutter, gutter 1, is equal to the distance of city B from gutter 3. On a particular day, the hospital in city A receives information that an accident has happened at gutter 3. The victim can be saved only if an operation is started within 40 min. An ambulance started from city A at 30 km/hr and crossed gutter 1 after 5 min. If the driver had doubled the speed after that, what is the maximum amount of time would the doctor get to attend the patient at the hospital. Assume 1 min is elapsed for taking the patient into and out of the ambulance?

• A. 4 min
• B. 2.5 min
• C. 1.5 min
• D. The patient died before reaching the hospital

Hint:

Ensure that all distances and times are calculated before finding the available time.

Answer 1

The Correct Option is C

- The ambulance is at city A at 30 km/h and crosses the first gutter after 5 minutes, so in 5 minutes the distance travelled is: \[ \text{Distance} = \frac{30}{60} \times 5 = 2.5 \, \text{km}. \] - The remaining distance to cover after doubling the speed to 60 km/h is: \[ \text{Distance left} = 20 \, \text{km} - 2.5 \, \text{km} = 17.5 \, \text{km}. \] - Time taken to cover the remaining distance at the doubled speed: \[ \text{Time} = \frac{17.5}{60} = 0.2917 \, \text{hrs} = 17.5 \, \text{minutes}. \] - So the total time spent is \( 5 \, \text{minutes} + 17.5 \, \text{minutes} = 22.5 \, \text{minutes}. \) - The time available is 40 minutes, and therefore the doctor gets \( 40 - 22.5 = 17.5 \) minutes for operation.